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# Balance the following chemical equations and identify the type of chemical reaction.${\mathrm{TiCl}}_{4}\left(\mathrm{l}\right)+\mathrm{Mg}\left(\mathrm{s}\right)\to \mathrm{Ti}\left(\mathrm{s}\right)+{\mathrm{MgCl}}_{2}\left(\mathrm{s}\right)$

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## Step 1: Given equation ${\mathrm{TiCl}}_{4}\left(\mathrm{l}\right)+\mathrm{Mg}\left(\mathrm{s}\right)\to \mathrm{Ti}\left(\mathrm{s}\right)+{\mathrm{MgCl}}_{2}\left(\mathrm{s}\right)$Step 2: Method used for balancing the equation Using the traditional method of balancing chemical equations. According to the method:First, compare the total number of atoms of each element on the reactant side and the product side.Add stoichiometric coefficients to molecules containing an element that has a different number of atoms on the reactant side and the product side.The coefficient must balance the number of atoms on each side.Usually, the stoichiometric coefficients for hydrogen and oxygen atoms are assigned last.The number of atoms of an element in one species must be obtained by multiplying the stoichiometric coefficient by the total number of atoms of that element present in 1 molecule of the species.Repeat this process till the number of all the atoms of the reacting elements are equal on both the reactant and product sides. Step 3: Balancing the given equationThe given equation is: ${\mathrm{TiCl}}_{4}\left(\mathrm{l}\right)+\mathrm{Mg}\left(\mathrm{s}\right)\to \mathrm{Ti}\left(\mathrm{s}\right)+{\mathrm{MgCl}}_{2}\left(\mathrm{s}\right)$Equation: ${\mathrm{TiCl}}_{4}\left(\mathrm{l}\right)+\mathrm{Mg}\left(\mathrm{s}\right)\to \mathrm{Ti}\left(\mathrm{s}\right)+{\mathrm{MgCl}}_{2}\left(\mathrm{s}\right)$Reactant SideProduct Side1 Titanium atom1 Titanium atom 4 Chlorine atoms 2 Chlorine atoms1 Magnesium atom1 Magnesium atomThe Titanium atoms are already balanced. Then the Chlorine atoms are balanced. After balancing the equation becomes: ${\mathrm{TiCl}}_{4}\left(\mathrm{l}\right)+\mathrm{Mg}\left(\mathrm{s}\right)\to \mathrm{Ti}\left(\mathrm{s}\right)+2{\mathrm{MgCl}}_{2}\left(\mathrm{s}\right)$Finally, the magnesium atoms are balanced. After balancing the equation becomes: ${\mathrm{TiCl}}_{4}\left(\mathrm{l}\right)+2\mathrm{Mg}\left(\mathrm{s}\right)\to \mathrm{Ti}\left(\mathrm{s}\right)+2{\mathrm{MgCl}}_{2}\left(\mathrm{s}\right)$The updated equation is: ${\mathrm{TiCl}}_{4}\left(\mathrm{l}\right)+2\mathrm{Mg}\left(\mathrm{s}\right)\to \mathrm{Ti}\left(\mathrm{s}\right)+2{\mathrm{MgCl}}_{2}\left(\mathrm{s}\right)$Equation: ${\mathrm{TiCl}}_{4}\left(\mathrm{l}\right)+2\mathrm{Mg}\left(\mathrm{s}\right)\to \mathrm{Ti}\left(\mathrm{s}\right)+2{\mathrm{MgCl}}_{2}\left(\mathrm{s}\right)$Reactant SideProduct Side1 Titanium atom1 Titanium atom 4 Chlorine atoms 4 Chlorine atoms 2 Magnesium atoms 2 Magnesium atomsSince the number of all the atoms of the reacting elements are equal on both the reactant and product sides are equal, the equation is balanced. Displacement Reaction A displacement reaction is a reaction in which the atom or a set of atoms is displaced by another atom in a molecule. In this reaction, the titanium atom is displaced by the magnesium atom. Hence this reaction is a displacement reaction. Hence the balanced equation is: ${\mathrm{TiCl}}_{4}\left(\mathrm{l}\right)+2\mathrm{Mg}\left(\mathrm{s}\right)\to \mathrm{Ti}\left(\mathrm{s}\right)+2{\mathrm{MgCl}}_{2}\left(\mathrm{s}\right)$ and this reaction is a displacement reaction.  Suggest Corrections  0      Similar questions
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